Problem 71
Question
The slope of the graph of \(y=x^{2}\) is different at every point on the graph of \(f\).
Step-by-Step Solution
Verified Answer
Yes, the slope of the function \(y = x^2\) is different at each point on its graph. The slope at a particular point is given by the derivative of the function which is \(2x\).
1Step 1: Understand the Function
The function given is \(y = x^2\), an elementary quadratic function. This type of function creates a parabola when graphed, which means the slope is not constant but changes at every point on the curve.
2Step 2: Find the Derivative of the Function
To find the slope of the graph at any given point, we need to compute the derivative of function \(y\). From calculus, we know that the derivative of \(x^2\) with respect to \(x\) is \(2x\). So, the derivative of function \(y\) i.e., \(y' = 2x\). This derivative gives us the slope of the function at any point \(x\).
3Step 3: Conclude the Analysis
The derivative of function \(y = x^2\), which represents the slope of the function at any given point, implies that indeed the slope is different depending on the value of \(x\). This validates that the slope of the graph of \(y = x^2\) is different at every point on the graph.
Key Concepts
Slope of a CurveQuadratic FunctionCalculus
Slope of a Curve
Understanding the **slope of a curve** is an essential part of calculus, especially when dealing with functions like quadratic equations. The slope tells us how steep a line is at a given point. For straight lines, the slope is constant. However, curves like those created by quadratic functions have slopes that change with different points along the curve.
For a quadratic function such as \( y = x^2 \), the slope at any point is not fixed. Instead, it depends on where you are on the curve. This variable slope is what makes understanding and calculating derivatives so important in calculus. By finding the derivative, we can determine the slope at any specific point on the curve. The derivative gives a mathematical formula to find these varying slopes. In the case of \( y = x^2 \), this derivative is \( y' = 2x \). This means that at any point \( x \), the slope of the curve is \( 2x \), indicating how fast \( y \) changes concerning \( x \) as we move along the curve.
For a quadratic function such as \( y = x^2 \), the slope at any point is not fixed. Instead, it depends on where you are on the curve. This variable slope is what makes understanding and calculating derivatives so important in calculus. By finding the derivative, we can determine the slope at any specific point on the curve. The derivative gives a mathematical formula to find these varying slopes. In the case of \( y = x^2 \), this derivative is \( y' = 2x \). This means that at any point \( x \), the slope of the curve is \( 2x \), indicating how fast \( y \) changes concerning \( x \) as we move along the curve.
Quadratic Function
**Quadratic functions** are a fundamental aspect of algebra and calculus. A quadratic function typically follows the form \( y = ax^2 + bx + c \). In our case, it takes the simple form \( y = x^2 \), which is the easiest quadratic function, where \( a = 1 \), and \( b \) and \( c \) are zero. These functions graph into a curve known as a parabola, which has a symmetrical shape that opens upwards or downwards depending on the sign of \( a \).
For our parabola \( y = x^2 \), it opens upwards, indicating that as \( x \) moves away from 0, \( y \) increases. The property of a parabola having a changing slope is because of its inherent curvature. The farther you get from its vertex (the lowest or highest point of the parabola), the steeper its slope becomes. Understanding quadratic functions is key as they appear in numerous real-world scenarios, from physics to economics, where they describe natural phenomena and trends.
For our parabola \( y = x^2 \), it opens upwards, indicating that as \( x \) moves away from 0, \( y \) increases. The property of a parabola having a changing slope is because of its inherent curvature. The farther you get from its vertex (the lowest or highest point of the parabola), the steeper its slope becomes. Understanding quadratic functions is key as they appear in numerous real-world scenarios, from physics to economics, where they describe natural phenomena and trends.
Calculus
**Calculus** provides the tools to understand and solve problems involving change and motion, and at its heart are derivatives. Derivatives help us find the slope of curves, such as those produced by quadratic functions. Calculus allows us to delve into analyzing rates of change systematically.
When faced with a quadratic function like \( y = x^2 \), calculus teaches us to find its derivative to understand how the curve behaves. By computing the derivative, we acquire a function \( y' = 2x \) which shows the rate of change or the slope at any given point \( x \).
When faced with a quadratic function like \( y = x^2 \), calculus teaches us to find its derivative to understand how the curve behaves. By computing the derivative, we acquire a function \( y' = 2x \) which shows the rate of change or the slope at any given point \( x \).
- **Differentiation** is what we refer to when finding this derivative, essentially dissecting a function into its rate of change components.
- Calculus also concerns itself with integrals, but primarily, derivative computations like this enable us to solve practical problems involving changing conditions.
Other exercises in this chapter
Problem 70
You deposit in an account that is compounded quarterly at an annual rate of \(r\) (in decimal form). The balance \(A\) after 10 years is \(A=2000\left(1+\frac{r
View solution Problem 71
Use the given information to find \(f^{\prime}(2)\) \(g(2)=3\) and \(g^{\prime}(2)=-2\) \(h(2)=-1 \quad\) and \(\quad h^{\prime}(2)=4\) $$ f(x)=g(x)+h(x) $$
View solution Problem 71
Consider a certificate of deposit that pays \(10 \%\) (annual percentage rate) on an initial deposit of \(\$ 1000\). The balance \(A\) after 10 years is \(A=100
View solution Problem 72
Use the given information to find \(f^{\prime}(2)\) \(g(2)=3\) and \(g^{\prime}(2)=-2\) \(h(2)=-1 \quad\) and \(\quad h^{\prime}(2)=4\) $$ f(x)=\frac{g(x)}{h(x)
View solution